将 class 类型作为函数参数传递并用作？施放 class

Question

有没有办法通过函数传递 class 类型并尝试将 class 转换为给定的 class 类型？我尝试了以下代码。

class Section {}
class TimeSection: Section {}
class TaskSection: Section {}

let timeSection = TimeSection()
let taskSection = TaskSection()

let sections = [timeSection, taskSection]

func findSection(from classType: Section.Type) {
    for section in sections {
        guard let section = section as? classType else { continue }

        print("Found section")
    }
}

findSection(from: TimeSection.self)

但我总是得到这个错误：

Use of undeclared type 'classType'

Answer 1

classType 实际上不是一个类型。它是一个包含 Section.Type 实例的参数。因此，您不能将它与 as?.

一起使用

既然是参数，可以和==比较一下。 ==的另一边应该是section的metatype的一个实例，可以通过type(of:)获取。

func findSection(from classType: Section.Type) {
    for section in sections {
        if type(of: section) == classType {
            print("Found section")
            break
        }
    }
}

Answer 2

Swift 4.2

您可以使用泛型函数并将类型参数限制为 Section。

import Foundation

class Section {}
class TimeSection: Section {}
class TaskSection: Section {}
class NoSection {}

let timeSection = TimeSection()
let taskSection = TaskSection()

let sections = [timeSection, taskSection]

func findSection<T: Section>(from classType: T.Type) {
    for section in sections {
        guard let section = section as? T else { continue }

        print("Found section: \(section)")
    }
}

findSection(from: TimeSection.self) // Found section: __lldb_expr_9.TimeSection
findSection(from: TaskSection.self) // Found section: __lldb_expr_9.TaskSection
findSection(from: NoSection.self) // won't compile

将 class 类型作为函数参数传递并用作？施放 class

Pass class type as a function parameter and use as? to cast the class

type-inference

swift

Swift 4.2